2016SDAU课程练习四1023 Problem W

Problem W
Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 48   Accepted Submission(s) : 19
[align=left]Problem Description[/align]
In 12th Zhejiang College Students Games 2007, there was a new stadium built in Zhejiang Normal University. It was a modern stadium which could hold thousands of people. The audience Seats made a circle. The total number of columns
were 300 numbered 1--300, counted clockwise, we assume the number of rows were infinite.<br>These days, Busoniya want to hold a large-scale theatrical performance in this stadium. There will be N people go there numbered 1--N. Busoniya has Reserved several
seats. To make it funny, he makes M requests for these seats: A B X, which means people numbered B must seat clockwise X distance from people numbered A. For example: A is in column 4th and X is 2, then B must in column 6th (6=4+2).<br>Now your task is to
judge weather the request is correct or not. The rule of your judgement is easy: when a new request has conflicts against the foregoing ones then we define it as incorrect, otherwise it is correct. Please find out all the incorrect requests and count them
as R.<br>
 

[align=left]Input[/align]
There are many test cases:<br>For every case: <br>The first line has two integer N(1<=N<=50,000), M(0<=M<=100,000),separated by a space.<br>Then M lines follow, each line has 3 integer A(1<=A<=N), B(1<=B<=N), X(0<=X<300) (A!=B), separated
by a space.<br><br>
 

[align=left]Output[/align]
For every case: <br>Output R, represents the number of incorrect request.<br>
 

[align=left]Sample Input[/align]

10 10
1 2 150
3 4 200
1 5 270
2 6 200
6 5 80
4 7 150
8 9 100
4 8 50
1 7 100
9 2 100

 

[align=left]Sample Output[/align]

2
<div style='font-family:Times New Roman;font-size:14px;background-color:F4FBFF;border:#B7CBFF 1px dashed;padding:6px'><div style='font-family:Arial;font-weight:bold;color:#7CA9ED;border-bottom:#B7CBFF 1px dashed'><i>Hint</i></div>
Hint:
（PS： the 5th and 10th requests are incorrect）
</div>

题目大意:

有n个人坐在zjnu体育馆里面，然后给出m个他们之间的距离， A B X， 代表B的座位比A多X. 然后求出这m个关系之间有多少个错误，所谓错误就是当前这个关系与之前的有冲突。

分析与总结：

题目有一句话：we assume the number of rows were infinite. 就是假设这个体育馆是无限大的，所以不用考虑圆圈循环的情况。

用带权并查集做， 对于并查集中的每一棵数， 树根的距离为0，然后以树根作为参照，每个结点的权值代表与树根的距离。

合并A，B时，假设A，B属于不同的树，那么就要合并这两棵树， 把A树合并到B树上，这时要给A树的跟结点root_a赋值，关键是给root_a附上一个什么值。 由于A点和B点的权值rank[A]和rank[B]都是相对跟结点的距离，所以分析A，B之间的相对距离，可以得到rank[root_a] = rank[A]+x-rank[B]。 注意到这时，对于原来的A的树，只跟新了root_a跟结点的权值， 那么其它结点的跟新在查找的那一步里面实行了。

#include<stdio.h>

 #include<string.h>

 #include<stdlib.h>

 #include<algorithm>

 #include<iostream>

 #include<queue>

 #include<map>

 #include<math.h>

 using namespace std;

 typedef long long ll;

 //typedef __int64 int64;

 const int maxn = 50015;

 const int inf = 0x7fffffff;

 const double pi=acos(-1.0);

 const double eps = 1e-8;

 

 int dis[ maxn ],fa[ maxn ];

 

 void init( int n ){

     for( int i=0;i<=n;i++ ){

         dis[ i ] = 0;

         fa[ i ] = i;

     }

     return ;

 }

 int find( int x ){

     if( fa[x]==x ) return x;

     int tmp = find( fa[x] );

     dis[ x ] = dis[ x ]+dis[ fa[x] ];

     dis[ x ] = dis[ x ]%300;

     fa[ x ] = tmp;

     return fa[ x ];

 }

 bool unionab( int a,int b,int c ){

     int x = find(a);

     int y = find(b);

     //printf("fa[%d]=%d,fa[%d]=%d,dis[%d]=%d,dis[%d]=%d\n",a,x,b,y,a,dis[a],b,dis[b]);

     if( x==y ){

         if( ((dis[b]-dis[a]+300)%300)!=c ){//b在a的后面，保证是dis[b]-dis[a]

             //printf("test:%d %d\n",a,b);

             return false;

         }

     }

     else{

         dis[ y ] = dis[ a ]+c-dis[ b ]+300;

         dis[y] %= 300;

         fa[ y ] = x;

     }

     return true;

 }

 int main(){

     int n,m;

     while( scanf("%d%d",&n,&m)==2 ){

         int a,b,c;

         init( n );

         int res = 0;

         while( m-- ){

             scanf("%d%d%d",&a,&b,&c);

             bool f = unionab( a,b,c );

             if( f==false ) res++;

         }

         printf("%d\n",res);

     }

     return 0;

 }